I've recently started studying Geometric Group Theory, and came across this interesting application of the Švarc-Milnor lemma to Riemannian Geometry.
Theorem) Let $M$ be a compact connected Riemann manifold without boundary, $\tilde{M}$ the universal cover of $M$. Then the fundamental group $\pi_{1}(M)$ is finitely generated and is quasi-isometric to $\tilde{M}$.
So that by the above theorem, if the fundamental group of a compact connected Riemann manifold is not quasi-isomorphic to $\mathbb{R}^{n}$ ($\mathbb{H}^n$), then it does not admit a flat (hyperbolic) Riemannian metric.
Many of the lecture notes that I've read so far includes this application of the lemma, but none gives a rigid proof (probably due to the fact that these lectures are not meant to dive heavily into Riemannian Geometry).
I only have some background in Differential Manifolds (not quite yet into Riemann Manifolds), some Algebraic Topology (familiar with the universal cover and deck transformation). So I decided to try and prove this theorem as an exercise, but failed. I've broken down the proof in the following steps:
- The universal cover $\tilde{M}$ of a Riemannian manifold $M$ (if there is one) is again a Riemannin manifold.
- In fact, $\tilde{M}$ is a proper geodesic metric space.
- The fundamental group $\pi_{1}(M)$ acts on $\tilde{M}$ via deck transformation, and this action is isometric.
- This action is properly discontinuous.
- This action is cobounded (or equivalently cocompact).
- Apply the Švarc-Milnor lemma.
Could someone help and explicitly write out the proof, or provide a reading material for each steps? I'd like to see an explicit proof and study it along with some necessary materials!
Here is the formal Švarc-Milnor lemma that I am looking at:
Lemma) Let $X$ be a proper geodesic metric space and let group $G$ act by isometry on $X$. If the action is properly discontinuous and cobounded, then $G$ is finitely generated, and $G$ is quasi-isometric to $X$.
Thanks in advance!